1. **Stating the problem:** We have three forces acting at a point: 100 N at an angle $\alpha = 33^\circ$ below the horizontal to the left, 180 N horizontally to the right, and 150 N at $30^\circ$ below the horizontal to the right. We want to find the resultant force vector. 2. **Formula and rules:** To find the resultant force, we sum the horizontal components and the vertical components of all forces separately: $$F_x = \sum F \cos \theta$$ $$F_y = \sum F \sin \theta$$ The magnitude of the resultant force is: $$F_R = \sqrt{F_x^2 + F_y^2}$$ The angle $\theta_R$ of the resultant force relative to the horizontal is: $$\theta_R = \tan^{-1}\left(\frac{F_y}{F_x}\right)$$ 3. **Calculate components:** - For 100 N force at $33^\circ$ left-downward: - Horizontal component: $100 \times \cos(180^\circ - 33^\circ) = 100 \times \cos 147^\circ$ - Vertical component: $100 \times \sin(180^\circ - 33^\circ) = 100 \times \sin 147^\circ$ - For 180 N force horizontally right: - Horizontal component: $180$ (positive) - Vertical component: $0$ - For 150 N force at $30^\circ$ right-downward: - Horizontal component: $150 \times \cos 30^\circ$ - Vertical component: $-150 \times \sin 30^\circ$ (negative because downward) 4. **Evaluate components:** - $\cos 147^\circ = \cos(180^\circ - 33^\circ) = -\cos 33^\circ = -0.8387$ - $\sin 147^\circ = \sin(180^\circ - 33^\circ) = \sin 33^\circ = 0.5446$ So, - $F_{x1} = 100 \times (-0.8387) = -83.87$ - $F_{y1} = 100 \times 0.5446 = 54.46$ - $F_{x2} = 180$ - $F_{y2} = 0$ - $F_{x3} = 150 \times \cos 30^\circ = 150 \times 0.8660 = 129.90$ - $F_{y3} = -150 \times \sin 30^\circ = -150 \times 0.5 = -75$ 5. **Sum components:** $$F_x = -83.87 + 180 + 129.90 = 226.03$$ $$F_y = 54.46 + 0 - 75 = -20.54$$ 6. **Calculate resultant magnitude:** $$F_R = \sqrt{226.03^2 + (-20.54)^2} = \sqrt{51074.76 + 421.88} = \sqrt{51496.64} = 226.99$$ 7. **Calculate resultant angle:** $$\theta_R = \tan^{-1}\left(\frac{-20.54}{226.03}\right) = \tan^{-1}(-0.0909) = -5.2^\circ$$ This means the resultant force is approximately $227$ N at $5.2^\circ$ below the horizontal to the right. **Final answer:** The resultant force has magnitude approximately $227$ N and direction $5.2^\circ$ below the horizontal to the right.